1. ## Linear equation problem

Hello everyone. I just don't get this . This seems very basic but I have problems to find a answer to this. I know augmented matrices and gaussian elimination.

Ok once again Ax = b
Okay I get the augmented matrix form. But how can I do this "backwards"

A is 2x2 matrix , and
x =[0;
3]
b = Null vector

if A was [a,b;c,d] then I came up with 3c + 3d = 0 and then set c =1 ,d=-1
This didn't function

At the moment I am incapable of coming up with a matrix which certain vectors are in the kernel or image. I can't make a matrix which needs to have certain solutions.

2. Hello,

If you mean:
$Ax=b$ where $
A=\begin{pmatrix}a&b\\c&d\end{pmatrix},\,
x=\begin{pmatrix}0\\3\end{pmatrix},\,
b=\begin{pmatrix}0\\0\end{pmatrix}$
,
you come up with:
$3b=0,\, 3d=0$.
Thus, $A=\begin{pmatrix}a&0\\c&0\end{pmatrix}$ where $a,\, c$ are arbitrary.

Bye.

3. Exactly what I meant.
$3b=0,\, 3d=0$. You got this by the multiplication $A=\begin{pmatrix}a&0\\c&0\end{pmatrix}$ * $x=\begin{pmatrix}0\\3\end{pmatrix}$ which should result in the 2x1 matrix
$3b=0,\, 3d=0$.

To solve those backwards , I take a*b and then rref it? Ok and that would then be your augmented matrix where for example 3b= say 4 if that was what I'd want

Ok I think I got this, thank you.